electrochemistry 3

E l e c t r o c h e m i s t r y C h 1 8 P a g e | 1

Electrochemistry Ch. 18 Notes:

Homework: Read Chapter 18,

Bonus: 39, 41, 43, 45, 47, 51, 55, 61, 63, 65, 67, 71, 73, 77, 81, 87, 89, 91, 99, 109

Check MasteringChemistry Deadlines

Electrochemistry:

All chemistry is electrical in the sense it involves electrons and other charged ions.

The term Electrochemistry is reserved for the study of the processes that convert chemical

energy to electrical energy and vice versa.

Spontaneous reactions are labeled Voltaic Cells or Galvanic Cells, derived from the

names Luigi Galvani (1737-1798) and Alessandro Volta (1745-1827). These cells use

spontaneous redox reactions to produce electric current, electrons flow from one terminal

(anode) to another (cathode) when connected by an external circuit and a salt bridge.

Example cases of spontaneous electrochemical processes include batteries, fuel cells and

corrosion.

Nonspontaneous reactions requiring electrical energy are called Electrolytic Cells;

electric current removes electrons from one reactant and gives electrons to another.

Example cases include electrolysis, used to decompose stable compounds such as H2O or

Fe2O3 into its elements, and electroplating.

For electrochemical reactions, a redox reaction can be separated in two, oxidation which

occurs at the anode and reduction that occurs at the cathode.

Review Redox Reactions: (found in chapter 4.9)

Redox reactions exchange electrons. When balancing a reaction, remember to

balance…

(1) the number of electrons lost = electrons gained

(2) the atoms on either side of the reaction must balance

(3) the total charge on either side of the reaction must balance.

Learn the half reaction method of balancing for electrochemistry reactions. This

method splits apart the oxidation half reaction from the reduction half reaction.


Oxidation Reduction increases oxidation number reduces oxidation number

lose electrons gain electrons

occurs at anode occurs at cathode

2 Cl-1

Cl2 + 2 e- K

+ + 1 e

- K

Oxidized substance, Cl-1

Reduced substance, K+

reductant/reducing agent oxidant/ oxidizing agent

Oil- Oxidation is losing e-‘s Rig- Reduction is gaining e

-‘s

Leo Ger

Assigning Oxidation Numbers

1) elements zero

2) monatomic ions charge on ion

3) oxygen mostly (-2) exceptions:

peroxide, O2-2

, (-1) and

superoxides, O2-1

, (-1/2)

4) hydrogen mostly (+1)

exceptions: hydrides (-1)

5) halogens mostly (-1)

Exceptions: (+1, +3, +5, +7)

when in a compound with lighter halogens or oxygen

6) Compounds sum of oxidation numbers equal zero

7) polyatomic ions sum of oxidation numbers equal charge of ion

Practice:

Determine the oxidation number of each element in the following list.

a) Cl2

b) MgCl2

c) Al(ClO)3

d) Ni(ClO2)2

e) V(ClO3)5

f) KClO4

g) K2Cr2O7

h) C2H4O

i) C3H8

Note: in most cases oxidation states are positive or negative integers. On occasion atoms within a

compound may have an average fractional oxidation state. It is acceptable because the oxidation states

are an imposed bookkeeping scheme, not an actual quantity.


Balancing Redox Rxns Using the Half Rxn Method

1) Assign oxidation numbers to each atom in the equation.

2) Determine what is oxidized and what is reduced.

3) Split the skeleton equation into two half reactions label appropriately as oxidation

or reduction reactions.

4) Complete and balance each half reaction

a) Balance atoms (except O and H) starting with the atom that changes

oxidation state.

b) Include electrons as products in the oxidation reaction and reactants in the

reduction reaction

c) If the solution is acidic: balance electric charges by adding H+1

ions, balance

the O’s by adding H2O, verify all atoms balance

d) If the solution is basic: balance electric charges by adding OH-1

ions, add

H2O to balance the O and H atoms, verify all atoms balance.

5) Combine the two half reactions to obtain overall equation

a) Multiply each half reaction by the factor needed to have equivalent electrons

lost and gained.

b) Simplify and reduce to the lowest whole number ratio

6) Check balancing

a) atoms

b) sum of charges of reactants equal to those of products

c) number on electrons lost and gained

Example 1: Balance the following with the half reaction method.

a) I2 (s) + S2O3-2

(aq) I-1

(aq) + S4O6-2

(aq)


b) A breathalyzer detects ethanol in suspected drunk drivers. Ethanol oxidizes to

acetaldehyde by dichromate ions in acidic solution. The orange dichromate turns

to green Cr+3

. Balance the reaction using the half-reaction method.

Cr2O7-2

(aq) + C2H5OH (l) Cr+3

(aq) + C2H4O (l) (acidic conditions)

c) I2 (s) + H2S (g) I-1

(aq) + S8 (s) (acidic conditions)

d) Al (s) + SO4-2

(aq) Al+3

(aq) + SO2 (g) (acidic conditions)


REDOX EQUATIONS: Answers to this additional practice found on website

Balance the following, identify species that oxidizes/reduces

1. Al + Pb+2

Al+3

+ Pb

2. Cu + NO3

-1 + H

+1 Cu

+2 + NO2 + H2O

3. Zn + NO3

-1 + H

+1 Zn

+2 + NH4

+1 + H2O

4. HI + HNO3 I2 + NO + H2O

5. OH-1

+ Cl2 Cl-1

+ ClO-1

+ H2O

6. Cr+3

+ MnO4

-1 Cr2O7

-2 + Mn

+2 (acidic)

7. MnO4

-1 + Mn

+2 MnO2 (acidic)

8. MnO4

-1 + IO3

-1 Mn

+2 + IO4

-1 (basic)

9. Br2 BrO3

-1 + Br

-1 (basic)


Voltaic Cells: also known as Galvanic Cells, Spontaneous:

Conduction through metal, such as wires, involves the flow of electrons and no

movement of atoms or obvious change in the metal.

Electrochemistry involves ionic conduction, the motion of ions through liquid,

anions spontaneously move toward the anode and cations spontaneous move toward

the cathode. The electrodes are connected externally by a metal wire in which

electrons move from the anode (-) to the cathode (+). Metal electrodes are often

part of the redox reaction. Inert electrodes do not react with reactants or products;

they just provide a surface for the reaction to occur. Common inert electrodes are

made of platinum metal or graphite (carbon).

In Voltaic Cells a spontaneous oxidation-reduction reaction produces electrical

energy if...

1) The two half cells are physically separated, internally by a salt bridge and

externally by a metal conduction wire.

2) Electron transfer is forced to occur along the external wire circuit.

3) A potential difference (voltage) is created making useful electrical energy.

Construction of Simple Galvanic Cell:

Half-cell: consists of oxidized and reduced forms of an element or more complex

species in contact with each other. Commonly, this is a metal electrode immersed in

a solution of its metal ions.

Two half cells are connected by a wire with a voltmeter inserted in the circuit to

measure the potential difference. Complete the circuit with a salt bridge.

Salt bridge: median through which ions slowly pass, filled with a nonreactive salt

and a substance to keep the salt solution from flowing out, this can be 5% agar

solution (gelatin from algae) or as in our lab, cotton yarn. A semipermeable porous

membrane often replaces the need for agar. The salt bridge serves 3 functions: 1)

makes electrical contact completing the circuit, 2) maintains electrical neutrality, 3)

prevents the spontaneous mixing of the half cells


Standard Cell: all reactants and products are in their standard states: pure liquid or

solid, 1M concentration of solutions, 1 atm pressure of gas, all at 25°C. A standard

cell is hard to maintain since the concentrations change as the reaction proceeds.

The initial voltage value gives us information on a standard cell. Slowly

equilibrium will be reached at which point the potential difference = 0 Volts.

Electromotive force = Emf = Ecell = Voltage, V = Joules/Coulomb (J/C).

1 electron has a charge of 1.60 x 10-19

C.

Ecell is positive for spontaneous reactions.

Shorthand Notation for Galvanic Cells: Drawing a diagram or using many

words and half reactions to describe the Galvanic cell becomes tedious.

Shorthand: Ni (s) | Ni +2

[0.180 M] || Cu+2

(? M) | Cu (s)

Anode | oxidation half cell || reduction half cell | Cathode.

The anode and cathode may be metals in the half reactions or inert electrodes.

Double line represents salt bridge. Vertical line is a phase barrier. If two species in

a half reaction are the same phase, comma may be used in place of vertical line.

Standard Galvanic Cells:

Zn (s)|Zn(NO3)2(aq) ||Cu(NO3)2 (aq)| Cu(s) E°cell= 1.10 V


Example 2: Examples of Standard Galvanic Cells:

Draw and label the following standard galvanic cells in the complete form:

a) Cu(s)|Cu(NO3)2(aq)||AgNO3(aq)|Ag(s) E°cell = 0.46 V

b) Pt (s) | Fe+2

(aq), Fe+3

(aq) || Cl2 (g)|Cl-1

(aq) | Pt (s) E°cell = 0.59 V


Standard Hydrogen Electrode and Standard Reduction Potentials:

An arbitrary zero standard reduction cell potential has been chosen by scientists to

be the SHE, Standard Hydrogen Electrode. In which the half cell has 1M H+1

, 1

atm H2 (g) and an inert Pt electrode.

2 H+ (aq, 1 M) + 2 e

-1 H2 (g, 1 atm) E°red = 0 V

Cell potentials :

A) Dependent on concentrations and based on SHE arbitrary voltage assignment

B) Never change when a reaction is doubled, tripled, or halved

C) Sign will reverse when the reaction is reversed

Zn+2

(aq, 1M) + 2 e-1

Zn (s) E°red = -0.76 V

2 Zn+2

(aq, 1M) + 4 e-1

2 Zn (s) E°red = -0.76 V no change

Zn (s) Zn+2

(aq, 1M) + 2 e-1

E°ox= +0.76 V reverse sign


Standard Reduction Cell Potential values are found in the appendix.

The Equation to solve for the standard cell potential of a reaction…

E°cell = E°red (cathode) – E°red (anode) or E°cell = E°ox + E°red

The cell potential must be positive to be spontaneous. That which is most easily

oxidized has the lower (greater negative) reduction potential. That which is most

easily reduced has the larger (greatest positive) reduction potential


Example 3:

Given that E°red Cu+2

= 0.34 V,

Write out the balanced half reactions and solve for the E°red for Zn+2

and Ag+1

a) Zn Zn(NO3)2 Cu(NO3)2 Cu E°cell= 1.10 V

b) Cu Cu(NO3)2 AgNO3 Ag E°cell = 0.46 V


Example 4:

Order the oxidation half reactions by increasing strength under standard conditions.

Half cells are as follows: Cu/Cu+2

, Zn/Zn+2

, H2/ H+

a) Write the balanced oxidation half reactions and standard half-cell potentials

for each.

b) Identify the species that is

1) the easiest to be oxidize

2) strongest reducing agent

3) strongest oxidizing agent

4) easiest to be reduced

Example 5:

Silver tarnish is mainly Ag2S solid coating silver metal objects

Ag2S (s) + 2e- 2 Ag (s) + S

-2 (aq) E°red = -0.691 V

A tarnished silver spoon is placed in contact with a commercially available metal

product in a glass-baking dish. Boiling water, to which some baking soda (salt

bridge) has been added, is poured into the dish, and the metal product and spoon are

completely covered. Within a short time, the spontaneous removal of tarnish from

the spoon begins.

Is this commercially available metal product Al or Cu? Explain by writing out the

oxidation half reaction and solve for the overall spontaneous E°cell value.


Free Energy and Redox Reactions:

Remember that Free Energy, G, is related to the equilibrium constant K.

G = G° + RTlnQeq and G° = -RTlnKeq

G° is also related to cell potential…

G° = -nFE°

Where n is the moles of electrons transferred and F is Faraday’s constant, the

electrical charge on one mole of electrons, F = 96,500 C/mol.

Remember that a Coulomb = J/V, E is in V, so G in J converted to kJ.

We can rearrange the equations above to relate cell potential to the equilibrium

constant and get…

nFE° = RTlnKeq and E°cell = ( ) lnKeq

An older form of the equation used logs under standard temperature

conditions (25°C)… E°cell = [(2.303 * 8.314J/mol K * 298K)/n(96485C)]log Keq

E°cell = (0.0592 Volts/n) log Keq

or E°cell = (0.0257/n) lnKeq at 25°C


Example 6:

Consider a reaction where tin (II) ions and solid tin is one half reaction and

cadmium ions and cadmium solid are the other half reaction in a spontaneous

voltaic cell under standard conditions and temperature.

a) Write out the shortened cell notation and determine E cell

b) Write out the overall reaction and determine the number of electrons transferred.

c) Calculate G°

d) Calculate, Keq

Cell EMF under Nonstandard Conditions, Nernst Equation:

Not all reactions occur under standard conditions. Even if a reaction starts with

standard conditions, over time the concentrations will change.

Manipulate the equations… G° = -nFE° and G = G° + RTlnQeq

to create the Nernst equation which allows us to calculate the cell potential

under all conditions.

Nernst equation: E = E° - (RT/nF)lnQ or E = E° - (2.303RT/nF)logQ

This allows us to calculate the electromotive force under nonstandard conditions.

Generally, the reaction is assumed at 25°C (298K), R = 8.314 J/mol K, and F =

96500J/V. Replacing the constants leads to…

E = E° - (0.0257/n)lnQ or E = E° - (0.0592/n)logQ (at 298K)


Example 7:

Electrochemical Cells can be used to determine unknown concentrations. This

may be useful in determining Ksp and molar solubility values or just any unknown

concentration in a balanced equation if others are known and the Ecell is

experimentally determined.

Given: Ecell = 1.075 V in the following reaction…

Cl2 (g) + Cu (s) Cu+2

(aq) + 2 Cl-1

(aq)

the anode mass is 15.0 g Cu (s),

Cl2 (g) partial pressure is 0.75 atm,

and 0.080 M Cu+2

(aq)

E°red Cu+2

/Cu = 0.34 V, E°red Cl2/Cl-1

= 1.36 V.

What is the concentration in molarity of Cl-1

ions in this problem?

Example 8:

The voltaic cell: Ni (s) | Ni+2 [0.180 M] || Cu+2 (? M) | Cu (s) has a cell

potential of 0.575V at 25°C. Diagram the voltaic cell. Use the Appendix to solve

for Ecell°. Solve for the concentration of copper ions.


Concentration Cell:

A concentration cell is a special case voltaic cell in which both half-cells have the

same species but different concentrations. The two cells will spontaneously react

when connected until equilibrium is reached. That equilibrium occurs when the

concentrations become equal and the Ecell = 0, this is the same equilibrium that

would be established if the two solutions are mixed directly. E°cell = 0 for a

concentration cell since E°ox and E°red are equal but opposite sign.

Ecell =0 - (0.0592/n)logQ

Example 9:

Solid zinc electrodes are in two half-reaction cells. One contains 0.500 M Zn+2

ions

and the other 0.0020 M Zn+2

ions. An exterior circuit and a salt bridge complete the

cells.

(a) Which concentration of Zn+2

ions will be reduced, (on the reactant side)?

(b) What is the initial spontaneous Ecell?

(c) What are the equilibrium concentrations of the Zn+2

ions in the cells?


Example 10:

A cell is constructed in which identical iron electrodes are placed in two solutions.

Solution A contains 0.880 M Fe(NO3)2. Solution B contains Fe+2

ions of a lower

concentration. A salt bridge and exterior wire circuit connect the two solutions.

The initial cell potential of the cell is observed to be 0.051 V.

a) Draw a diagram of the cell described above and label well.

b) Which solution, A or B, is changing the iron ions to solid iron?

c) What is the initial [Fe+2

] in solution B?

d) What is Ecell and the concentration of [Fe+2

] in each of the solutions when the

system reaches equilibrium?


Batteries and Fuel Cells: Types include Primary, Secondary and Fuel Cells…

Primary Galvanic Cells are irreversible. These are common store bought batteries.

The patent in 1866 for a Leclanche Cell also

known as a Dry Cell has the anode as a zinc

outer casing,

Zn Zn+2

+ 2e- E°ox 0.76V

inside is a moist paste separated by a paper

acting as a salt bridge

the cathode is inert graphite in the center

touching moist NH4Cl and MnO2

Reduction cell reaction is

2NH4+ + 2e

-2 NH3+H2 E°red 0.84V

Ecell = Eox + Ered = 1.60 V(decreases as

the cell is used)

MnO2 is needed to prevent build up of H2.

Zn+2

reacts with the NH3 to form a complex

ion. Side reactions eliminate products to

prevent a decrease in cell potential.

H2 + 2 MnO2 Mn2O3 (s) + H2O (l)

Zn+2

+ 4 NH3 Zn(NH3)4+2

Alkaline Dry cell batteries are similar to

the1866 patent except a more basic

environment (KOH) is used giving the battery

a longer shelf life.

Modern batteries last longer under heavy use.


Secondary Galvanic Cells are rechargeable. During discharge the forward reactions

occur and when direct current passes through the cell to recharge the reverse reactions

occur.

Lead acid storage

batteries used in cars

batteries are a typical

example.

The following reactions occur during discharge...

Anode: Pb Pb+2

+ 2 e- E°ox = 0.356 V

Pb+2

+ SO4-2

PbSO4

Cathode: PbO2 + 4 H+ + 2 e

- Pb

+2 + H2O E°red = 1.685 V

Pb+2

+ SO4-2

PbSO4

Overall: Pb + PbO2 + 2 H2SO4 2 PbSO4 + H2O E°cell = 2.041 V

This is set up in a series of 6 cells in order to make a 12V battery.

Other common rechargeable batteries include Nickel-Cadmium (nicad). Nickel-

Metal Hydride, Lithium and Lithium Ion batteries.

Example 11:

Nickel-cadmium batteries are more expensive than alkaline batteries but they

are light, rechargeable, and can be sealed to prevent leakage.

The batteries in many calculators and in cordless tools are nickel-cadmium

batteries.

In a NiCad battery the unbalanced half reactions are Cd(s) Cd(OH)2(s) and

NiO2 (s) Ni(OH)2 (s), which take place in basic solution.

a) Balance the half reactions in basic solution.

Reaction at anode

Cd (s) Cd(OH)2 (s) E°ox = 0.81 V

Reaction at cathode

NiO2 (s) Ni(OH)2 (s) E°red = 0.49 V


b) Write the overall reaction and include the E°cell under standard conditions for

the nickel-cadmium battery as it spontaneously discharges?

c) In some batteries the power continually decreases toward zero volts in cell

potential as the reactants are used up and products formed, in others the cell

voltage remains fairly constant during discharge. Explain the reason for the

difference and determine what is expected for the NiCad battery above.

Fuel Cells have a continuous supply of reactants and removal of product and are

50-70% efficient. A H2 and O2 fuel cell that creates water has been used in space

crafts for power for decades. Now fuel cell batteries are being made to fuel alternative

vehicles, equipment, buildings and some day residential homes.

Anode: (H2 + 2 OH-1

2 H2O + 2 e-) x 2

Cathode: O2 + 2 H2O + 4 e- 4 OH

-

Overall: 2 H2 + O2 2 H2O

with catalyst Ag, Pt, CoO

Newer fuel cells replaced the strong OH-1

electrolyte with a special proton

exchange membrane (PEM).

The PEM is a plastic membrane that conducts protons, not electrons.

Anode: 2H2 4 H+1

+ 4 e-

Cathode: O2 + 4 H+1

+ 4 e- 4 H2O

Overall: 2 H2 + O2 2 H2O


Corrosion:

Corrosion is an unwanted problem associated with spontaneous electrochemical

reactions. Corrosion is the spontaneous oxidation of a metal by chemicals in the

environment, mainly O2.

Because many materials we use are active metals, corrosion can be a very big

problem

Metals are often used for their strength and malleability, but these properties

are lost when the metal corrodes.

For many metals, the product of corrosion also does not adhere to the metal,

and as it flakes off more metal can corrode.

Rusting:

• At the anodic regions, Fe(s) is oxidized to Fe2+

• The electrons travel through the metal to a cathodic region where O2 is reduced

in acidic solution from gases dissolved in the moisture

• The Fe2+

ions migrate through the moisture to the cathodic region where they are

further oxidized to Fe3+

which combines with the oxygen and water to form rust

rust is hydrated iron(III) oxide, Fe2O3•nH2O

the exact composition depends on the conditions

moisture must be present

water is a reactant

required for ion flow between cathodic and anodic regions

• Electrolytes promote rusting

enhances current flow

• Acids promote rusting

lowering pH will lower E°red of O2

A great deal of money is spent every year trying to prevent or replace corrosion

problems. Five ways of reducing corrosion follows.


Reducing and Preventing Corrosion.

a) Plating a metal with a thin layer of less easily oxidized metal. Example: tin can

to store food. Tin coated iron works since Sn Sn+2

+ 2e- has E°ox = 0.14 V

and is less easily oxidized compared to iron, Fe Fe+2

+ 2e- has E°ox = 0.44 V

Problem: If the container is dented and the iron layer is exposed, corrosion

occurs faster than an unprotected iron container by the new electrochemical cell

made with Sn and Fe.

b) Connecting the metal directly to a "sacrificial anode" which is more easily

oxidized. Example: underground iron pipe, bridges, or ships. When Zn is

attached to Fe the Zn is preferentially oxidized protection the Fe since Zn

Zn+2

+ 2e- has E°ox = 0.763 V which is higher than Fe.

Problem: Over time the sacrificial anode corrodes and will need to be replaced.

c) Allow a protective film to naturally form on metal surface. Example: 1) the

green copper patina on the statue of liberty in which Cu exposed to the

environment forms CuCO3.Cu(OH)2 2) Al reacts to form Al2O3. In both cases

a tough hard inert surface forms over the metal and prevents further damage.

Problem: Acid rain will dissolve the protective coatings that have been

established.


d) Galvanizing, or coating an object with a

more easily oxidized metal. This is

similar to the "sacrificial anode"; only the

entire object is coated. Example:

galvanized nails in which the Fe nail is

coated with a thin layer of Zn. The iron

stays protected and strong even when bent

and exposed since the Zn will oxidize

preferentially.

e) Apply a protective coating over the metal.

Example: paint, wax or a sealant.

Example 11:

Give a minimum of three useful ways to protect an ocean-going ship from

corrosion. Identify the pro’s and con’s. Which methods are most beneficial?


Electrolysis or Electrolytic Cells, Nonspontaneous:

Electrolytic Cells are nonspontaneous and need power to react. Ecell is a

negative value. Aluminum was once a very expensive metal to refine since it was

originally isolated by electrolysis methods. No salt bridge is necessary since the

reaction will not spontaneously occur. You must separate the products to avoid

them spontaneously returning to the reactants. Anode (+) and cathode (−)

Examples of electrolysis:

Aqueous NaI produces I2 (s) and H2 (g)

possible oxidations

2 I− I2 + 2 e

− E°ox = −0.54 V

2H2O O2 +4e−+4H

+ E°ox = −0.82 V

possible reductions

Na+ + 1e

− Na

0 E°red = −2.71 V

2H2O+ 2e−

H2 + 2OH

− E°red = −0.41 V

overall reaction

2 I−

(aq) + 2 H2O(l) I2(aq) + H2(g) + 2 OH−

(aq) E°cell = −0.95 V

That which is easier to accomplish will occur.

When more than one cation is present, the cation that is easiest to reduce will be reduced

first at the cathode

least negative or most positive E°red

When more than one anion is present, the anion that is easiest to oxidize will be oxidized

first at the anode

least negative or most positive E°ox


Molten NaCl (l) produces Cl2 gas and solid Na

Electrolysis of pure compounds:

• The compound must be in molten (liquid) state

• Electrodes normally graphite

• Cations are reduced at the cathode to metal element

• Anions oxidized at anode to nonmetal element

Aqueous NaCl (aq) produces H2 and Cl2 gases

Write out the possible reactions and predict the expected products…

The actual reaction requires a greater voltage for the oxidation of water to oxygen than the predicted

0.82V at pH = 7 (overvoltage occurs). The overvoltage increases the value to about 1.4V causing Cl2 to

oxidize more easily.


Electrolysis of water in Na2SO4 (aq) solution produces O2 and H2 gases

possible oxidations

2 SO4−2 S2O8

-2 + 2 e

− E°ox = −2.01 V

2H2O O2 +4e−+4H

+ E°ox = −0.82 V

possible reductions

Na+ + 1e

− Na

0 E°red = −2.71 V

2H2O+ 2e−

H2 + 2OH

− E°red = −0.41 V

Predict the overall reaction:

Electrolysis of Aqueous Solutions:

• Possible cathode reactions

reduction of cation to metal

reduction of water to H2

2 H2O + 2 e−

H2 + 2 OH

− E° = −0.83 v @ stand. cond.

E° = −0.41 v @ pH 7

• Possible anode reactions

oxidation of anion to element

oxidation of H2O to O2

2 H2O O2 + 4 e− + 4H

+ E° = −1.23 v @ stand. cond.

E° = −0.82 v @ pH 7

• Half-reactions that lead to least negative Ecell will generally occur unless

overvoltage changes the conditions

Overvoltage often occurs for the oxidation of water to O2 in aqueous

solutions due to the large number of electrons and H+ ions involved as well

as other kinetic factors. With the overvoltage the oxidation of water requires

about 1.4V. The underlying kinetic factors that cause this are beyond the

scope of this class.


Applications of Electrolysis:

Electrolytic Refining is the most practical, but expensive method to isolate active

metals like Na. Chemical reactions can separate many less active metals from their

ores. Usually further refining is still needed for purification. Example: the anode

contains impure copper and the cathode is purified Cu. As power is forced through the

system the pure Cu thickens and grows and the impure Cu thins and drops off a sludge

of impurities to the bottom of the vessel.

Electroplating of Metals uses the same method as the refining by electroplating a thin

layer of a metal such as Ag or Au on jewelry.

In electroplating, the work

piece is the cathode

Cations are reduced at cathode

and plate to the surface of the

work piece

The anode is made of the plate

metal. The anode oxidizes and

replaces the metal cations in

the solution.


Quantitative Aspects of Electrolysis:

Volt = Joule/Coulomb

Current is the number of electrons that flow through the system per second, Ampere

Coulomb = Amp . sec (C = A

. s) and Faraday's constant = 96500 C/mole e

-

Electrical work is generally measured in watts. 1 W = 1 J/s.

One kilowatt hour (kWh) = (1000W) (3600s/hr) = 3.6 x 106 J

Example 12:

An unknown aqueous iron salt is electrolyzed by a current of 2.75 A for 45 minutes

producing 1.43 g of solid Fe. Solve for the oxidation number for Fe in the salt?

Example 13:

Solve for the grams of Cr (s) that will be deposited from CrO4-2

(aq) using 6.00

amps current for 3 1/2 hours. Assume the theoretical yield of Cr (s) is obtained.

Example 14:

How many kWh are required to produce 5.00 kg of Mg from the electrolysis of

molten MgCl2 if the applied cell potential is 5.00 V. Assume 100% efficiency.

electrochemistry 3

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oxidation reduction

oxidation half reaction

es rig reduction

number of electrons

assign oxidation numbers

cases oxidation states

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